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A particle performs simple harmonic moti...

A particle performs simple harmonic motion with amplitude A. Its speed is tripled at the instant that it is at a distance
`(2A)/3` from equilibrium position. The new amplitude of the motion is:

A

`(7A)/(3)`

B

`(A)/(3)sqrt(41)`

C

`3A`

D

`sqrt(3)A`

Text Solution

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The correct Answer is:
A
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